Problem: This is the easy version of the problem. The only difference is that here k=2. You can make hacks only if both the versions of the problem are solved.

This is an interactive problem.

Every decimal number has a base k equivalent. The individual digits of a base k number are called k-its. Let's define the k-itwise XOR of two k-its a and b as (a + b)mod k.

The k-itwise XOR of two base k numbers is equal to the new number formed by taking the k-itwise XOR of their corresponding k-its. The k-itwise XOR of two decimal numbers a and b is denoted by a⊕_{k} b and is equal to the decimal representation of the k-itwise XOR of the base k representations of a and b. All further numbers used in the statement below are in decimal unless specified. When k = 2 (it is always true in this version), the k-itwise XOR is the same as the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

You have hacked the criminal database of Rockport Police Department (RPD), also known as the Rap Sheet. But in order to access it, you require a password. You don't know it, but you are quite sure that it lies between 0 and n-1 inclusive. So, you have decided to guess it. Luckily, you can try at most n times without being blocked by the system. But the system is adaptive. Each time you make an incorrect guess, it changes the password. Specifically, if the password before the guess was x, and you guess a different number y, then the system changes the password to a number z such that x⊕_{k} z=y. Guess the password and break into the system.

Input

The first line of input contains a single integer t (1≤ t≤ 10 000) denoting the number of test cases. t test cases follow.

The first line of each test case contains two integers n (1≤ n≤ 2⋅ 10^5) and k (k=2).

It is guaranteed that the sum of n over all test cases does not exceed 2⋅ 10^5.

Interaction

For each test case, first read two integers n and k. Then you may ask up to n queries.

For each query, print a single integer y (0≤ y≤ 2⋅ 10^7). Let the current password be x. After that, read an integer r.

If x=y, you will read r=1 and the test case is solved. You must then continue solving the remaining test cases.

Else, you will read r=0. At this moment the password is changed to a number z such that x⊕_{k} z=y.

After printing a query, do not forget to output the end of line and flush the output. Otherwise, you will get the Idleness limit exceeded verdict.

To do this, use:

  * fflush(stdout) or cout.flush() in C++; 
  * System.out.flush() in Java; 
  * flush(output) in Pascal; 
  * stdout.flush() in Python; 
  * see documentation for other languages. 



If you ask an invalid query or exceed n queries, you will read r=-1 and you will receive the Wrong Answer verdict. Make sure to exit immediately to avoid unexpected verdicts.

Note that the interactor is adaptive. That is, the original password is not fixed in the beginning and may depend on your queries. But it is guaranteed that at any moment there is at least one initial password such that all the answers to the queries are consistent.

Hacks:

To use hacks, use the following format of tests:

The first line should contain a single integer t (1≤ t≤ 10 000) — the number of test cases.

The first and only line of each test case should contain two integers n (1≤ n≤ 2⋅ 10^5) and k (k=2) denoting the number of queries and the base respectively. The optimal original password is automatically decided by the adaptive interactor.

You must ensure that the sum of n over all test cases does not exceed 2⋅ 10^5.

Example

Input


1
5 2

0

0

1


Output


3

4

5

Note

In the example test case, the hidden password is 2.

The first query is 3. It is not equal to the current password. So, 0 is returned, and the password is changed to 1 since 2⊕_2 1=3.

The second query is 4. It is not equal to the current password. So, 0 is returned, and the password is changed to 5 since 1⊕_2 5=4.

The third query is 5. It is equal to the current password. So, 1 is returned, and the job is done.

Note that this initial password is taken just for the sake of explanation. When you submit, the interactor might behave differently because it is adaptive.

Solution (in JAVA):

Solution:


import java.util.*;
import java.io.*;

public class RPD_and_Rap_sheet {
	static StringBuilder sb;
	static dsu dsu;
	static long fact[];
	static long mod = (int) (1e9);

	static void solve() {
		int n = i();
		int k = i();

		int i = 0;
		for (int j = 0; j < n; j++) {
			int curr = i ^ j;
			System.out.println(curr);
			System.out.flush();
			int temp = i();
			if (temp == 1) {
				return;
			}
			i = i ^ curr;
		}
	}

	public static void main(String[] args) {
		sb = new StringBuilder();
		int test = 1;
		test = i();
		while (test-- > 0) {
			solve();
		}
		// System.out.println(sb);

	}

	/*
	 * fact=new long[(int)1e6+10]; fact[0]=fact[1]=1; for(int i=2;i<fact.length;i++)
	 * { fact[i]=((long)(i%mod)1L(long)(fact[i-1]%mod))%mod; }
	 */
//**************NCR%P******************	 
	static long ncr(int n, int r) {
		if (r > n)
			return (long) 0;

		long res = fact[n] % mod;
		// System.out.println(res);
		res = ((long) (res % mod) * (long) (p(fact[r], mod - 2) % mod)) % mod;
		res = ((long) (res % mod) * (long) (p(fact[n - r], mod - 2) % mod)) % mod;
		// System.out.println(res);
		return res;

	}

	static long p(long x, long y)// POWER FXN //
	{
		if (y == 0)
			return 1;

		long res = 1;
		while (y > 0) {
			if (y % 2 == 1) {
				res = (res * x) % mod;
				y--;
			}

			x = (x * x) % mod;
			y = y / 2;

		}
		return res;
	}

//**************END******************

	// *************Disjoint set
	// union*********//
	static class dsu {
		int parent[];

		dsu(int n) {
			parent = new int[n];
			for (int i = 0; i < n; i++)
				parent[i] = -1;
		}

		int find(int a) {
			if (parent[a] < 0)
				return a;
			else {
				int x = find(parent[a]);
				parent[a] = x;
				return x;
			}
		}

		void merge(int a, int b) {
			a = find(a);
			b = find(b);
			if (a == b)
				return;
			parent[b] = a;
		}
	}

//**************PRIME FACTORIZE **********************************//
	static TreeMap<Integer, Integer> prime(long n) {
		TreeMap<Integer, Integer> h = new TreeMap<>();
		long num = n;
		for (int i = 2; i <= Math.sqrt(num); i++) {
			if (n % i == 0) {
				int nt = 0;
				while (n % i == 0) {
					n = n / i;
					nt++;
				}
				h.put(i, nt);
			}
		}
		if (n != 1)
			h.put((int) n, 1);
		return h;

	}

//****CLASS PAIR ************************************************
	static class Pair implements Comparable<Pair> {
		int x;
		int y;

		Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}

		public int compareTo(Pair o) {
			return (int) (this.y - o.y);

		}

	}
//****CLASS PAIR **************************************************

	static class InputReader {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;

		public InputReader(InputStream stream) {
			this.stream = stream;
		}

		public int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public int Int() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public String String() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public boolean isSpaceChar(int c) {
			if (filter != null)
				return filter.isSpaceChar(c);
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public String next() {
			return String();
		}

		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}

	static class OutputWriter {
		private final PrintWriter writer;

		public OutputWriter(OutputStream outputStream) {
			writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
		}

		public OutputWriter(Writer writer) {
			this.writer = new PrintWriter(writer);
		}

		public void print(Object... objects) {
			for (int i = 0; i < objects.length; i++) {
				if (i != 0)
					writer.print(' ');
				writer.print(objects[i]);
			}
		}

		public void printLine(Object... objects) {
			print(objects);
			writer.println();
		}

		public void close() {
			writer.close();
		}

		public void flush() {
			writer.flush();
		}
	}

	static InputReader in = new InputReader(System.in);
	static OutputWriter out = new OutputWriter(System.out);

	public static long[] sort(long[] a2) {
		int n = a2.length;
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a2)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < l.size(); i++)
			a2[i] = l.get(i);
		return a2;
	}

	public static int[] sort(int[] a2) {
		int n = a2.length;
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a2)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < l.size(); i++)
			a2[i] = l.get(i);
		return a2;
	}

	public static long pow(long x, long y) {
		long res = 1;
		while (y > 0) {
			if (y % 2 != 0) {
				res = (res * x);// % modulus;
				y--;

			}
			x = (x * x);// % modulus;
			y = y / 2;
		}
		return res;
	}

//GCD___+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
	public static long gcd(long x, long y) {
		if (x == 0)
			return y;
		else
			return gcd(y % x, x);
	}
	// ******LOWEST COMMON MULTIPLE
	// *********************************************

	public static long lcm(long x, long y) {
		return (x * (y / gcd(x, y)));
	}

//INPUT PATTERN********************************************************
	public static int i() {
		return in.Int();
	}

	public static long l() {
		String s = in.String();
		return Long.parseLong(s);
	}

	public static String s() {
		return in.String();
	}

	public static int[] readArrayi(int n) {
		int A[] = new int[n];
		for (int i = 0; i < n; i++) {
			A[i] = i();
		}
		return A;
	}

	public static long[] readArray(long n) {
		long A[] = new long[(int) n];
		for (int i = 0; i < n; i++) {
			A[i] = l();
		}
		return A;
	}

}